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love history [14]
2 years ago
10

A certain simple pendulum has a period on the earth of 2.00 s . part a what is its period on the surface of mars, where g=3.71m/

s2?
Physics
1 answer:
Orlov [11]2 years ago
6 0
You have to use the period length equation for a pendulum to get the length of the pendulum then plug the length in with the new gravity to get 3.25 seconds.
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Jack tries to place magnets on the door of his refrigerator. He observes that the magnets don’t stick. He guesses that the door
zaharov [31]
The answer is D.) Make observations
3 0
2 years ago
Read 2 more answers
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a
Pie

Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

v = \sqrt{\frac{GM}{R+h}}

here we know that

M = 5.98 \times 10^{24} kg

R = 6.37 \times 10^6 m

h = 900 km = 9.0 \times 10^5 m

now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

v = 7407.1 m/s

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

v_{rel} = \sqrt{2} v

v_{rel} = 1.05 \times 10^4 m/s

6 0
2 years ago
What is the mass, in grams, of 2.00 moles of H2O?
tia_tia [17]

Answer:

36g

Explanation:

Given parameters:

Number of moles of H₂O = 2moles

Unknown:

Mass of  H₂O = ?

Solution:

To solve this problem, use the expression below:

   Mass of  H₂O = number of moles x molar mass

Molar mass of  H₂O = 2(1) + 16  = 18g/mol

  Mass of  H₂O = 2 x 18  = 36g

3 0
2 years ago
Help me it’s for a test I have??
Volgvan
1. C
2. A
3. E
4. D
5. B
6. F
i might have 2 and 6 mixed up, not completely sure tho
4 0
2 years ago
Read 2 more answers
A 2.8-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts
agasfer [191]

Answer with Explanation:

We are given that

Mass of one cart,m_1=2.8 kg

Mass of second cart,m_2=1.2 kg

Initial velocity of one cart,u_1=4.6m/s

Initial velocity of second cart,u_2=-2.7 m/s

a.Total momentum,P=m_1u_1+m_2u_2=2.8(4.6)+1.2(-2.7)

P=9.64 kgm/s

b.Velocity of second cart,v_2=0

According to law of conservation of momentum

Initial momentum=Final momentum

9.64=2.8v_1+1.2\times 0

v_1=\frac{9.64}{2.8}

v_1=3.44m/s

8 0
3 years ago
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